Misc 35 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Misc 35 Prove that β«_0^(π/2)βsin^3β‘π₯ ππ₯=2/3 Solving L.H.S β«_0^(π/2)βsin^3β‘π₯ ππ₯ = β«_0^(π/2)βγ γsin π₯ (sinγ^2β‘γπ₯)γ γ ππ₯ = β«_0^(π/2)βπ ππβ‘γπ₯ (1βγπππ γ^2 π₯)γ ππ₯ = β«_0^(π/2)βπ ππβ‘γπ₯ ππ₯β β«1_0^(π/2)βγsinβ‘γπ₯ γπππ γ^2 π₯γ ππ₯γγ π°_π β«1_0^(π/2)βsinβ‘γπ₯ ππ₯γ = β [cosβ‘π₯ ]_0^(π/2) = β[0β1] = 1 π°_π β«1_0^(π/2)βsinβ‘γπ₯ γπππ γ^2 π₯ ππ₯γ Let t = cos x ππ‘/ππ₯ = - sin x dt = β sin x dx Substituting, β«1_0^1βsinβ‘γπ₯Γπ‘^2Γγ ππ‘/(βsinβ‘γπ₯ γ ) = ββ«1_1^0βγπ‘^2 ππ‘γ = γβ[π‘^3/3]γ_1^0 = β("0 β " 1/3)=β((β1)/3) = 1/3 L.H.S = πΌ_1β πΌ_2 = 1 β 1/3 = π/π = R.H.S Hence, proved.
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo